Question 160437


{{{4x^2+5x-2=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=5}}}, and {{{c=-2}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(5) +- sqrt( (5)^2-4(4)(-2) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=5}}}, and {{{c=-2}}}



{{{x = (-5 +- sqrt( 25-4(4)(-2) ))/(2(4))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--32 ))/(2(4))}}} Multiply {{{4(4)(-2)}}} to get {{{-32}}}



{{{x = (-5 +- sqrt( 25+32 ))/(2(4))}}} Rewrite {{{sqrt(25--32)}}} as {{{sqrt(25+32)}}}



{{{x = (-5 +- sqrt( 57 ))/(2(4))}}} Add {{{25}}} to {{{32}}} to get {{{57}}}



{{{x = (-5 +- sqrt( 57 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-5+sqrt(57))/(8)}}} or {{{x = (-5-sqrt(57))/(8)}}} Break up the expression.  



So the answers are {{{x = (-5+sqrt(57))/(8)}}} or {{{x = (-5-sqrt(57))/(8)}}} 



which approximate to {{{x=0.319}}} or {{{x=-1.569}}}